Acceleration as a function of t

Classical Mechanics Level 1

Suppose that the displacement $s$ of an aeroplane can be expressed as a function of time $t$ in the following way: $s(t) = 8 t^2 + 10 t .$ What is the magnitude of the acceleration of the object?

32 \text{ m/s}^{2}

8 \text{ m/s}^{2}

16 \text{ m/s}^{2}

4 \text{ m/s}^{2}

4 solutions

Sukhpreet Saini
Jun 2, 2014

$x=8t^{2}+10t\\ a=\dfrac{dv}{dt}$

Also, $v=\dfrac{dx}{dt}\\ \dfrac{dx}{dt}=16x+10\\ v=16x+10$

Therefore $a=16m/s^{2}$

could u plz explain it in a simpler way cause I am getting it at all

jahnavi g - 6 years, 10 months ago

first at all if you dont get it they used defrentiation aka taking the derivative of the function on t (time) if you don't know what a derivative is google it or maybe find a wiki here on it but simply what they are doing is a = v/t v=x/t(x here is distance)

so they calculated the rate of change(derivative) of distance(x) as a function of time to get the rate of change of x (aka x/t or v) then having V they did the same thing calculating the rate of change (derivative) of velocity as a function of time (aka v/t) getting in the end A=v/t which is acceleration oh and the symbol d next to the x and t just means (a small change in ) which we use bec we are dealing with graphs and taking a slope of a graph needs deffrentiation from where the d comes from(x/t or dx/dt is the slop s(t) at any point)

Oximas omar - 1 year, 6 months ago

Emmanuel David
Oct 30, 2017

$Displacement: s(t) = 8t^2 + 10t$

$Velocity: \frac{d}{dt} (s(t)) = 16t + 10$

$Acceleration: \frac{d}{dt}(\frac{d}{dt} (s(t))) = \boxed{16}$

Yokme Estoque
Nov 30, 2014

X=8t^{2}+10t a=dv/dt Also v=dx/dt dx/dt=16x+10 V=16x+10 Therefore a=16m/s^{2}

Supathat Sukaiem
Jun 7, 2014

Haha.. I just recall the formulas.

½a = 8

Thus, a = 16.

Acceleration as a function of t

4 solutions

0 pending reports